Gears & gear trains
- Meshing gears
- n1·N1 = n2·N2
- Torque
- T2 = T1·(N2/N1)·η
- Power
- P = T·ω, ω = 2π·n/60
- Compound train
- total ratio = product of the stage ratios
Machine Design: cram pack
Formula sheets, the key method steps for every topic and 12 practice problems, with full solutions at the back.
1Formulas and methods
Torsion of shafts
- Shear stress
- τ = T·r/J
- Solid shaft
- J = πd⁴/32
- Hollow shaft
- J = π(D⁴ − d⁴)/32
- Angle of twist
- φ = TL/(GJ)
- Power
- P = T·ω = 2πnT/60
Method
- Find the polar moment of area: J = πd⁴/32 for a solid shaft.
- Shear stress at the surface: τ = T·r/J, with r the outer radius.
- Angle of twist: φ = TL/(GJ) in radians.
- If power is given, find the torque first with P = T·ω, ω = 2πn/60.
Mohr’s circle & stress transformation
- Centre
- C = (σx + σy)/2
- Radius
- R = √(((σx − σy)/2)² + τxy²)
- Principal stresses
- σ1,2 = C ± R
- Principal direction
- tan 2θp = 2τxy/(σx − σy)
- Inclined plane
- σθ = C + (σx − σy)/2·cos 2θ + τxy·sin 2θ
Method
- Calculate the centre of Mohr’s circle: C = (σx + σy)/2.
- Calculate the radius: R = √(((σx − σy)/2)² + τxy²).
- The principal stresses are C ± R and the largest in-plane shear stress is R.
- Angles on the circle are double the angles on the element: tan 2θp = 2τxy/(σx − σy).
Axial stress & strain
- Normal stress
- σ = F/A
- Strain
- ε = ΔL/L
- Hooke’s law
- σ = E·ε
- Elongation
- ΔL = FL/(EA)
- Thermal strain
- εT = α·ΔT
Column buckling
- Euler load
- Pcr = π²EI/(KL)²
- Effective length factor
- K = 1 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2 fixed–free
- Critical stress
- σcr = Pcr/A
- Slenderness
- λ = KL/r, r = √(I/A)
Mechanical vibrations
- Natural frequency
- ωn = √(k/m), fn = ωn/(2π)
- From static deflection
- ωn = √(g/δst)
- Critical damping
- cc = 2√(km) = 2mωn
- Damping ratio
- ζ = c/cc
- Damped frequency
- ωd = ωn√(1 − ζ²)
- Logarithmic decrement
- δ = (1/n)·ln(x0/xn), ζ = δ/√(4π² + δ²)
- Springs in parallel
- k = k1 + k2
- Springs in series
- 1/k = 1/k1 + 1/k2
2Practice problems
1.Gears & gear trainsMedium
A 1.5 kW motor at 1450 rpm drives a worm with 4 start(s). The worm wheel has 72 teeth and the drive has an efficiency of 80%. Find the output torque Tw on the wheel shaft.
z = 4Nw = 72n = 1450 rpmP = 1.5 kWη = 80%
Tw = N·m
2.Gears & gear trainsHard
An open belt connects a 150 mm pulley and a 500 mm pulley whose centres are 500 mm apart. Find the angle of wrap θ on the smaller pulley, in degrees.
d1 = 150 mmd2 = 500 mmC = 500 mm
θ = °
3.Torsion of shaftsMedium
A solid brass shaft 1.2 m long carries a torque of 1000 N·m. The angle of twist between its ends must not exceed 0.5°. Find the smallest diameter d that meets this limit.
T = 1000 N·mL = 1.2 mG = 39 GPaφallow = 0.5°
d = mm
4.Torsion of shaftsHard
A solid aluminium shaft of diameter 50 mm is fixed at A. A torque of 1500 N·m acts at B, 1.5 m from A, and a torque of 750 N·m in the opposite direction acts at the free end C, a further 0.8 m along. Find the maximum shear stress τmax in the shaft.
d = 50 mmTB = 1500 N·mTC = 750 N·mAB = 1.5 mBC = 0.8 m
τmax = MPa
5.Mohr’s circle & stress transformationMedium
A point in a loaded part has σₓ = 160 MPa, σᵧ = −30 MPa and τₓᵧ = −60 MPa (tension positive). Find the angle θp from the x-axis to the direction of the largest principal stress, between −90° and 90°.
σx = 160 MPaσy = −30 MPaτxy = −60 MPa
θp = °
6.Mohr’s circle & stress transformationHard
A point in a loaded part has σₓ = 150 MPa, σᵧ = 30 MPa and τₓᵧ = 10 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 15° counter-clockwise from the x-axis.
σx = 150 MPaσy = 30 MPaτxy = 10 MPaθ = 15°
σθ = MPa
7.Axial stress & strainMedium
A bar is made of a steel segment (diameter 40 mm, length 600 mm, E = 200 GPa) joined end to end with a segment made of copper (diameter 40 mm, length 400 mm, E = 120 GPa). It is fixed at one end and pulled at the other. Find the axial load P that stretches it by 0.3 mm in total.
segment 1: steel, Ø40 mm, 600 mm, E = 200 GPasegment 2: copper, Ø40 mm, 400 mm, E = 120 GPaδ = 0.3 mm
P = kN
8.Axial stress & strainHard
A steel rod (diameter 40 mm, E = 200 GPa, allowable stress 200 MPa) sits inside a tube made of brass (outer diameter 100 mm, inner diameter 70 mm, E = 100 GPa, allowable stress 70 MPa). They are compressed together between rigid plates. Find the largest load Pmax.
rod: Ø40 mm, E = 200 GPa, σallow = 200 MPatube: Ø100/Ø70 mm, E = 100 GPa, σallow = 70 MPa
Pmax = kN
9.Column bucklingMedium
A column of length L made of steel is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
L = 2 mE = 200 GPa (steel)d = 60 mm
Pcr = kN
10.Column bucklingHard
A solid round column of length L made of aluminium is fixed at the base and free at the top. It must carry 115 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
P = 115 kNn = 2.5L = 3 mE = 70 GPa (aluminium)
d = mm
11.Mechanical vibrationsMedium
In a free-vibration test of a damped mass–spring system, a displacement peak of 40 mm is followed 5 cycles later by a peak of 1.7 mm. Find the damping ratio ζ.
x0 = 40 mmx5 = 1.7 mmn = 5
ζ =
12.Mechanical vibrationsHard
A 10 kg mass hangs from spring k₃ = 25 kN/m. The top of k₃ is attached to a rigid bar that hangs from two springs side by side, k₁ = 20 kN/m and k₂ = 5 kN/m; the bar stays level. Ignore damping and the masses of the springs and bar. Find the natural frequency fn of vertical vibration in Hz.
m = 10 kgk1 = 20 kN/mk2 = 5 kN/mk3 = 25 kN/m
fn = Hz
3Solutions
1.Gears & gear trains
Answer: Tw = 142.3 N·m
Gear trains: meshing gears have the same speed at the teeth, so n₁·N₁ = n₂·N₂: the gear with fewer teeth turns faster. Gears on one shaft turn at the same speed. Power is passed on (minus losses), so torque goes up when speed goes down: T₂ = T₁·(N₂/N₁)·η, and P = T·ω with ω = 2π·n/60.
Wheel speed
nw = n·z/Nw = 1450·4/72 = 80.56 rpm
Output torque
ωw = 2π·80.56/60 = 8.436 rad/s
Tw = η·P/ωw = 0.8·1500/8.436 = 142.3 N·m
Answer
Tw = 142.3 N·m
2.Gears & gear trains
Answer: θ = 139 °
Belt drives: without slip, both pulleys have the same rim speed v = π·d·n/60, so n₁·d₁ = n₂·d₂ and the speed ratio is d₂/d₁. Slip s makes the driven pulley a little slower: n₂ = n₁·(d₁/d₂)·(1 − s). The power transmitted is P = (T₁ − T₂)·v, with T₁ the tension on the tight side and T₂ on the slack side. Friction limits the ratio of the tensions: T₁/T₂ = eμθ, where θ is the angle of wrap in radians on the smaller pulley; for an open belt θ = π − 2·asin((d₂ − d₁)/(2C)) with centre distance C.
Belt angle
asin(0.35) = 20.49°
Angle of wrap
θ = 180 − 2·20.49 = 139°
Answer
θ = 139°
3.Torsion of shafts
Answer: d = 77.41 mm
Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.
Required polar moment
T = 1000 N·m = 1 × 10⁶ N·mm
L = 1.2 m = 1200 mm
G = 39 GPa = 39000 MPa
φ = 0.5·π/180 = 0.008727 rad
J = TL/(Gφ) = 1 × 10⁶·1200/(39000·0.008727) = 3.526 × 10⁶ mm⁴
Diameter
d = (32J/π)1/4 = 77.41 mm
Answer
d = 77.41 mm
In practice you would round up to the next standard size.
4.Torsion of shafts
Answer: τmax = 30.56 MPa
Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.
Internal torques
Sense of TC taken as positive.
TBC = TC = 750 N·m
TAB = TC − TB = 750 − 1500 = −750 N·m
Maximum shear stress
The segment with the largest internal torque has the largest stress (same diameter).
|T|max = 750 N·m
τmax = 16·750000/(π·50³) = 30.56 MPa
Answer
τmax = 30.56 MPa
5.Mohr’s circle & stress transformation
Answer: θp = −16.14 °
Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.
Angle on the circle
Point x sits at an angle 2θp from the σ-axis on Mohr’s circle; the quadrant follows from the signs of σₓ − σᵧ and τₓᵧ.
2θp = atan2(2·−60, 160 − −30) = −32.28°
Angle on the element
θp = −16.14°
Answer
θp = −16.14°
Positive angles are counter-clockwise.
6.Mohr’s circle & stress transformation
Answer: σθ = 147 MPa
Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.
Transformation equation
σθ = 90 + 60·0.866 + 10·0.5 = 147 MPa
Answer
σθ = 147 MPa
7.Axial stress & strain
Answer: P = 59.52 kN
Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, Pi = P·EiAi/ΣEA, and σi = Ei·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.
Areas
A₁ = 1257 mm², A₂ = 1257 mm²
Flexibilities
L₁/(E₁A₁) = 600/(200000·1257) = 2.387 × 10⁻⁶ mm/N
L₂/(E₂A₂) = 400/(120000·1257) = 2.653 × 10⁻⁶ mm/N
Load
P = 0.3/5.04 × 10⁻⁶ = 59525 N = 59.52 kN
Answer
P = 59.52 kN
8.Axial stress & strain
Answer: Pmax = 456.3 kN
Two materials: in series (one after the other) every segment carries the same force and the elongations add: δ = P·Σ L/(E·A). In parallel (side by side between rigid plates) both parts shorten by the same amount, so their strains are equal: the stiffer part takes the larger share, Pi = P·EiAi/ΣEA, and σi = Ei·ε. A bar held between two rigid walls with a load at the junction is statically indeterminate: the two segments share the load in proportion to their stiffness k = EA/L.
Areas
Ar = π·40²/4 = 1257 mm²
At = π(100² − 70²)/4 = 4006 mm²
Combined axial stiffness
ΣEA = 200·10³·1257 + 100·10³·4006 = 6.519 × 10⁸ N
Governing strain
ε = min(0.001, 0.0007) = 0.0007 (brass)
Largest load
Pmax = 0.0007·6.519 × 10⁸ = 456316 N = 456.3 kN
Answer
Pmax = 456.3 kN
9.Column buckling
Answer: Pcr = 640.7 kN
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
Smallest second moment of area
A round section has the same I about every axis.
I = πd⁴/64 = π·60⁴/64 = 636173 mm⁴
Effective length
KL = 0.7·2000 = 1400 mm
Critical load
Pcr = π²EI/(KL)² = π²·200000·636173/1400² = 640691 N = 640.7 kN
Answer
Pcr = 640.7 kN
10.Column buckling
Answer: d = 132.2 mm
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
Required critical load
Pcr = n·P = 2.5·115000 = 287500 N
KL = 2·3000 = 6000 mm
Solve for the diameter
d = (64·n·P·(KL)²/(π³E))^(1/4) = (64·287500·6000²/(π³·70000))^(1/4) = 132.2 mm
Answer
d = 132.2 mm
Check: the slenderness KL/r is 181.6, so the column is slender and Euler applies.
11.Mechanical vibrations
Answer: ζ = 0.1
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Logarithmic decrement
δ = (1/5)·ln(40/1.7) = 0.6317
Damping ratio
ζ = δ/√(4π² + δ²) = 0.6317/√(39.48 + 0.399) = 0.1
Answer
ζ = 0.1
12.Mechanical vibrations
Answer: fn = 5.627 Hz
Free vibration of a mass–spring system: the undamped natural frequency is ωn = √(k/m) in rad/s, and fn = ωn/(2π) in Hz. A viscous damper c gives the damping ratio ζ = c/cc, with critical damping cc = 2·√(k·m) = 2·m·ωn. An underdamped system (ζ < 1) oscillates at ωd = ωn·√(1 − ζ²). Springs in parallel add, k = k₁ + k₂; springs in series add as reciprocals, 1/k = 1/k₁ + 1/k₂.
Equivalent stiffness
Only k1 and k2 are in parallel. That pair is in series with k3, so combine them as reciprocals.
k12 = 20 + 5 = 25 kN/m
keq = 25·25/(25 + 25) = 12.5 kN/m
Natural frequency
ωn = √(12500/10) = 35.36 rad/s
fn = 35.36/(2π) = 5.627 Hz
Answer
fn = 5.627 Hz
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